similar to: fraction [a] a partitioning of variation

Dear R-users I applied vegan's varpart function to partition the effects of explanatory matrices. Adj. R square for the unique fraction [a] is 0.25. Does anyone know why the decomposition by hand using rda gives me a different result for [a] (constrained proportion is 0.32)? I used cbind() for the conditional fractions, but it should be similar to condition()? Thanks very much

Hello I am using Jari Oksanen's CCA routine from the Vegan package on some estuary data, following a technique applied in (Anderson, M.J. & Gribble, N.A., 1998, Partitioning the variation among spatial, temporal and environmental components in a multivariate data set, Australian Journal of Ecology 23, 158-167). Some steps in the process require that the dependent matrix be constrained by

Hi, I did a linear regression with 5 explanatory variables. Now, to see the contribution of each variable, I use varpart from vegan library. But varpart don?t accepts my 5 explanatory variables, it accept only 4. 1- How must I do to use my 5 explanatory variables? 2- Is it the sum of variance fraction of each variable must be equal to 1. Thanks for your help. Komine -- View this

Hello, I currently run aov in the following way: > throughput.aov <- aov(log(Throughput)~No_databases+Partitioning+No_middlewares+Queue_size,data=throughput) > summary(throughput.aov) Df Sum Sq Mean Sq F value Pr(>F) No_databases 1 184.68 184.675 136.6945 < 2.2e-16 *** Partitioning 1 70.16 70.161 51.9321 2.516e-12 *** No_middlewares 2 44.22

I have some data (from Affymetrix experiments) where I fit an aov() model to a large number of outcome variables. A reasonable fraction of the outcome variables have 0 variability because values below a cutoff have been replaced with the cutoff value (often 20) . In this case, the overall p-value from summary(lm(..)) is misleadingly small: For example (equivalent results in R 1.6.1 on both

I have been trying to draw histogram for my manscript and found some strange things that I could not figure out why. Using the same code listed below I have successfully draw histograms for a few figures with fraction labeled on Y axis less than 1 (acturally between 0 to 0.1). But one dataset gives the Y axis label 0 to 5 as fraction. This is not true, as fraction are less than 1, although the

Hi, Dear Greg, Sorry to bother you again. I have several questions about the 'gbm' package. if the train.fraction is less than 1 (ie. 0.5) , then the* first* 50% will be used to fit the model, the other 50% can be used to estimate the performance. if bag.fraction is 0.5, then gbm use the* random* 50% of the data to fit the model, and the other 50% data is used to estimate the

Hi R-Users, I have a student doing work with lionfish and she has been trying to analyse a multivariate dataset to see what variables/factors are influencing the behaviour of lionfish. We have attempted a number of analyses, including rpart, relimpo and standard linear regression but we are not having much luck with quality output. The data is very non-normal and we would appreciate some advice

Hello: Has anyone tried to model-based recursive partition (using mob from package party; thanks Achim and colleagues) a data set based on a multinomial logit model (using mlogit from package mlogit; thanks Yves)? I attempted to do so, but there are at least two reasons why I could not. First, in mob I am not quite sure that a model of class StatModel exists for mlogit models. Second, as

Dear R users I wonder if there is a quick way to calculate the ratio/fraction of a list/data frame. For example, if I have a data frame with two fields: "Index" and "A". I would like to know the fractions of A's within the same "Index". That is, for Index =1, three fractions will be "1/(1+2+3)=0.17", "2/(1+2+3)=0.33", and

subject: trimmed mean in R seems to round the trimming fraction to r-help at stat.math.ethz.ch. Consider the following example of 10 numbers. 10% trimmed mean is correct but you can see that the result is the same for many trimming fractions till 0.20! For example 13% trimmed mean should use interpolation of second and eighth ordered observation. R does not seem to do this. The correct 13%

Hi all, This is really a stats question as much as an R question. I'm trying to do a joint scaling test (JST - see below) on some very oddly-distributed data and was wondering if anyone can suggest a good way of dealing with model violations and/or using R to evaluate how sensitive the model is to violations of the normality assumption. Here's a quick explanation of the analysis, the

Dear r-help, I'm having this DF df <- data.frame(id = 1:6, xout = c(1234, 2134, 234, 456, 324, 345), xin= c(NA, 34,67,87,34, NA)) and would like to calculate the fraction (xin_t / xout_t-1) The result should be: # NA, 2.76, 3.14, 37.18, 7.46, NA I am sure there is a solution using zoo... but I don't know how... Thanks for any help! Patrick

Dear list, I understand that to raise matrix A to power (-1/2) we should use something like this: eigen(A)$vectors%*%diag(1/sqrt(eigen(A)$values))%*%t(eigen(A)$vectors) [from previous discussions: http://r.789695.n4.nabble.com/matrix-power-td900335.html] But this will only do it for negative sqrt of the matrix not for other fraction powers like (-3/2). Seeing that these things be can done

I have this code: Link.find(params[:id]).toggle!(''active'') Before it, its timestamp is: 2005-11-20 20:45:48.741973-07 After: 2005-11-20 20:45:48-07 Why does Rails take it upon itself to truncate the seconds fraction? How can I get Rails to not change it? Thanks, csn __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best

Hello I like to call to 2 providers provider X = IAX provider Y = SIP exten => _06.,1,Dial(IAX2/X/${EXTEN},30,r)&(SIP/${EXTEN}@Y.Y.Y.Y) exten => _06.,2,Hangup Provider X is working but provider Y never shows up. What's wrong ?? How can I get provider Y working a fraction earlier the provider X Thanks Sjaak

I have used packages rpart, mvpart and tree for classification and regression trees. I want to calculate fraction of null deviance explained by each node and variable in the tree. For instance, at the first split, this would be (1 - (sum of residual deviance in each of the two leaves)/deviance at the root). In the subsequent splits, this formula is slightly different. There probably is a function

Hi all, I would like to plot part of the emperical CDF. Suppose the variable is x, I just need the part when x>1,therefore, I am using the following codes. tail <- x>1 plot(ecdf(x[tail]), do.points=FALSE, verticals=TRUE) The "x" value starts from 1, but the yaxs still begins from 0, not the corresponding value when "x" is 1. How can I make it match? Could anyone

I am trying to find the important factors from a two level fraction factorial design (2^{8-4}) I have studied the companion of vikneswaran but would really like a basic example of how to do this: (adapted to 8 factors): mydata<-read.table("myfile.txt", header=T) summary(aov(resp ~ A*B*C*D*E*F*G*H, data = mydata)) However i can't figure out how the input files are supposed to

Hi all R users I did a logistic regression with my binary variable Y (0/1) and 2 explanatory variables. Now I try to draw my nomogram with predictive value. I visited the help of R but I have problem to understand well the example. When I use glm fonction, I have a problem, thus I use lrm. My code is: modele<-lrm(Y~L+P,data=donnee) fun<- function(x) plogis(x-modele$coef[1]+modele$coef[2])